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kotegsom [21]
3 years ago
12

riangle XYZ is translated 4 units up and 3 units left to yield ΔX'Y'Z'. What is the distance between any two corresponding point

s on ΔXYZ and ΔX'Y'Z′?
Mathematics
1 answer:
tester [92]3 years ago
7 0

Answer:

5 units

Step-by-step explanation:

According to the given statement  Δ XYZ is translated 4 units up and 3 units left to yield ΔX'Y'Z' which means that each point in ΔXYZ is moved 4 units up and moved 3 units left.

To find the distance of each corresponding point we will use the Pythagorean theorem which states that the square of the length of the Pythagorean of a right triangle is equal to the sum of the squares of the length of other legs

The square of the required distance = 4^2+3^2 = 16+9 =25

By taking root of 25 we get:

√25 = 5

Thus, we can conclude that the the distance between any two corresponding points on ΔXYZ and ΔX′Y′Z′  is 5 units. ..

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What is the distance between M(9, −5) and N(−11, 10)?
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Answer:

\boxed{ \bold{ \huge{ \boxed{ \sf{25 \: units}}}}}

Step-by-step explanation:

Let M ( 9 , -5 ) be ( x₁ , y₁ ) and N ( - 11 , 10 ) be ( x₂ , y₂ )

<u>Finding</u><u> </u><u>the </u><u>distance </u><u>between</u><u> </u><u>these</u><u> </u><u>points</u>

\boxed{ \sf{distance =  \sqrt{ {(x2 - x1)}^{2} +  {(y2 - y1)}^{2}  } }}

\longrightarrow{ \sf{ \sqrt{ {( - 11 - 9)}^{2}  +  {(10 - ( - 5))}^{2} } }}

\longrightarrow{ \sf{ \sqrt{ {( - 20)}^{2} +  {(10 + 5)}^{2}  } }}

\longrightarrow{ \sf{ \sqrt{ {( - 20)}^{2}  +  {(15)}^{2} } }}

\longrightarrow{ \sf{ \sqrt{400 + 225}}}

\longrightarrow{ \sf{ \sqrt{625}}}

\longrightarrow{ \sf{ \sqrt{ {(25)}^{2} } }}

\longrightarrow{ \sf{25 \: units}}

Hope I helped!

Best regards! :D

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